Défi :

Étant donné un mot, vérifiez s'il s'agit ou non d'un isogramme.

Quelle :

Un isogramme est un mot composé uniquement de lettres sans doublons (insensible à la casse). La chaîne vide est un isogramme.

Exemples :

"Dermatoglyphics"     ---> true
"ab"                  ---> true
"aba"                 ---> false
"moOse"               ---> false
"abc1"                ---> false
""                    ---> true

Contribution :

Vous pouvez accepter une entrée dans n'importe quel format raisonnable

L'entrée ne contiendra que des lettres et / ou des chiffres, pas d'espaces ( [a-zA-Z0-9])

Production :

• true ou toute valeur vraie si l'entrée est un isogramme
• false ou toute valeur de fausse autrement

C'est le code-golf donc le code le plus court en octets dans chaque langue gagne.

Python deux / 3, 36 52 48 octets

lambda s:len(s)==len({*s.lower()}-{*str(56**7)})

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Je profite du fait qu'il setne contient que des éléments uniques. En invoquant la __len__méthode de chacun, je peux déterminer si scontient également uniquement des éléments uniques (en ignorant la casse).

EDIT: mis à jour pour satisfaire l'exigence précédemment ignorée de renvoyer False pour les entrées numériques. L'ensemble de tous les chiffres est codé comme set(str(56**7)).

EDIT 2: Suite à cette suggestion d'utilisateur , je profite maintenant du déballage des arguments pour définir la compréhension. Cela rompt formellement la compatibilité avec Python 2.

05AB1E , 5 octets

lDáÙQ

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Explication

l        # convert input to lower-case
 D       # duplicate
  á      # keep only letters
   Ù      # remove duplicates
    Q     # compare for equality

R , 41 octets

!grepl("(.).*\\1|\\d",tolower(scan(,"")))

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Approche regex. !grepl(regex,scan(,""),F)n'a pas fonctionné, donc je suppose que la capture ne correspond pas à la casse dans R? Je suis mauvais en regex en général, cependant, je ne serai donc pas surpris si je fais juste mal ...

R , 58 octets

!anyDuplicated(c(el(strsplit(tolower(scan(,"")),"")),0:9))

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Ajoute les chiffres 0:9à la liste (en minuscules) de caractères et teste s'il y a des doublons.

Rubis , 25 23 21 octets

-2 octets sur les deux grâce à Giuseppe

->s{/(.).*\1|\d/i!~s}

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-2 octets grâce à Kirill L.

Rubis -n , 21 19 18 16 octets

p !/(.).*\1|\d/i

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Brachylog , 4 octets

ḷo⊆Ạ

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Le prédicat réussira si l'entrée est un isogramme et échouera dans le cas contraire, produisant l'alphabet latin en minuscules s'il réussit. Étant donné que le ⊆prédicat intégré de Brachylog ne correspond pas exactement à la relation ordinaire entre un sous-ensemble et un sur-ensemble, j'ai dû dépenser un octet pour trier l'entrée en minuscules, mais j'ai économisé un octet pour ne pas avoir à vérifier explicitement les doublons. (Si cela n'avait pas besoin d'échouer avec les chiffres, nous pourrions simplement l'utiliser ḷ≠.)

Husk , 6 octets

§=f√ü_

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Explication

§=f√ü_  -- takes a string as argument, eg: "sAad2"
§       -- fork the argument..
  f√    -- | filter out non-letters: "sad"
    ü_  -- | deduplicate by lower-case: "sAd2"
 =      -- ..and compare: 0

Japt , 12 octets

;v
oC ‰ eUq

Explication:

;v
;         // Set alternative default vars, where C is the lowercase alphabet
 v        // Make the implicit input lowercase and reassign it
oC ‰ eUq
oC        // Remove all items from the input that are not in the alphabet
   ‰     // Split into chars and select unique array items
      eUq // Check if the result is equal to the input split into chars

Essayez-le ici.

MATL , 9 octets

kt2Y2X&X=

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k   % Lowercase implicit input
t   % Duplicate that
2Y2 % Push lowercase alphabet
X&  % Intersection of alphabet and duplicate lowercase input
X=  % Check for exact equality.

Python 3 , 46 octets

lambda s:s.isalpha()*len(s)==len({*s.lower()})

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Japt 2.0, 12 11 bytes

-1 byte thanks to Nit

v
f\l â eUq

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JavaScript (Node.js), 29 25 bytes

s=>!/(.).*\1|\d/i.test(s)

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Thanks for the update on answer to @BMO , @l4m2 , @KevinCruijssen

-4 bytes thanks to @KevinCruijssen

Retina, 16 bytes

Ci`(.).*\1|\d
^0

Returns 1 as Truthy and 0 as Falsey values.
Thanks @Neil for discovering and fixing a bug in my initial code.

Try it online.

Explanation:

C             Check if the input matches part of the following regex:
 i`            Case insensitivity enabled
               Check if part of the input matches either:
  (.)           A character `C`
     .*         followed by zero or more characters
       \1       followed by the same character `C` again
         |     Or
          \d    A digit
^0             Invert Truthy/Falsey, basically replacing every 0 with a 1,
               and every other value with a 1

PowerShell, 91 bytes

param($b)($a=[char[]]$b.ToUpper()|group|sort c*)[0].Count-eq$a[-1].count-and$b-notmatch'\d'

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Naive solution, but I can't come up with a better algorithm. Takes input $b, converts it ToUppercase, casts it as a char-array. Pipes that array into Group-Object which constructs a object that has name/count pairs for each input letter. We then sort that based on the count and take the 0th one thereof. We check that its .Count is -equal to the .Count of the last [-1] pair. If so, then the counts are all equal, otherwise we have a different amount of letters.

We then -and that with checking whether the input -notmatches against \d to rule out any digits in the input. That Boolean result is left on the pipeline and output is implicit.

Jelly, 8 bytes

ŒufØA⁼QƲ

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Python 2, 57 56 bytes

x=input().lower()
print len(set(x)-set(`763**4`))/len(x)

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First it turn then input into a set, removing the duplicates, then remove the digits (encoded in `763**4`), then check if the length is the same as the original input

Java 8, 61 39 bytes

s->!s.matches("(?i).*((.).*\\2|\\d).*")

Explanation:

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s->  // Method with String parameter and boolean return-type
  !s.matches("(?i).*((.).*\\2|\\d).*")
     //  Return whether the input does not match the regex

Regex explanation:

String#matches implicitly adds ^...$.

^(?i).*((.).*\2|\d).*$
 (?i)                      Enable case insensitivity
^    .*                    Zero or more leading characters
       (       |  )        Followed by either:
        (.)                 Any character `C`
           .*               with zero or more characters in between
             \2             followed by that same character `C` again
               |           Or:
                \d          Any digit
                   .*$     Followed by zero or more trailing characters

APL (Dyalog Unicode), 12 bytes

Anonymous tacit function.

(∪≡~∘⎕D)819⌶

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819⌶ lowercase

(…) apply the following tacit function on that:

 ~∘⎕D remove Digits from the argument

 ∪≡ are the unique elements of the argument identical to that?

Perl 6, 22 bytes

{!(.uc~~/(.).*$0|\d/)}

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No matches for some character then later the same character. Implicit function as a code block, match implicitly on $_, invert book with !. Added |\d (ta Adam) but also needed .uc~~, which needed parentheses...

Alternative with Bags, 23 bytes

{.uc.ords.Bag⊆65..97}

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This one normalises case then makes a bag (set with incidence counts). Subset or equal only true if all members are members of the comparison Bag, and all incidence counts are less than or equal to those in the comparison Bag. So any repeats or digits would make the comparison false.

Swift, 81 bytes

let s=readLine()!.lowercased().characters;print(s.count<1||Set(s).count==s.count)

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Visual Basic for Applications (32 bit), 102 bytes

s=LCase(InputBox(u)):j=1:For i=1To Len(s):k=Mid(s,i,1):j=j*0^Instr(i+1,s,k)*(k Like"[a-z]"):Next:?j<>0

Used the fact that in VBA 0^x yields 1 if x is zero and 0 otherwise. Run in immediate (debug) window.

Edit: as pointed out by Taylor in the comments this only works in 32 bit installs of MS Office.

05AB1E, 4 bytes

lDÔQ

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Explanation

l      # convert input to lowercase
 D     # duplicate and push to stack
  Ô    # uniquify the list of characters
   Q   # check for equality

C (gcc), 87 85 83 bytes

• Saved two four bytes thanks to ceilingcat.

f(s,p,b,P)char*s,*p;{for(b=s;*s;++s)for(p=b*=*s>64;b&&p<s;b=(*s^*p++)&95?b:0);s=b;}

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K (ngn/k), 18 bytes

{(a^,/$!10)~?a:_x}

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CJam, 11 bytes

qelA,s+_L|=

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Explanation

The basic idea is to append each digit then check for duplicates. Since the append ensures that each digit is already present once, any further presence of digits will be a duplicate, causing it to return false.

q      e# read the input:            | "MOoSE1"
el     e# convert to lowercase:      | "moose1"
A      e# push 10:                   | "moose1" 10
,      e# range [0,N):               | "moose1" [0 1 2 3 4 5 6 7 8 9]
s      e# string representation:     | "moose1" "0123456789"
+      e# concatenate:               | "moose10123456789"
_      e# duplicate:                 | "moose10123456789" "moose10123456789"
L|     e# union with the empty list: | "moose10123456789" "mose1023456789"
       e# (this gets rid of duplicates)
=      e# Equal to original:         | 0

Red, 76 bytes

func[s][a: charset[#"a"-#"z"#"A"-#"Z"]parse s[any[copy c a ahead not to c]]]

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Smalltalk, 57 bytes

Method to be defined in class String:

s^(self select:#isLetter)asUppercase asSet size=self size

This is most likely self-explanatory.

Pyth, 17 bytes

.Am&!t/rz0d}dGrz0

Test suite

.Am&!t/rz0d}dGrz0 # Code
  m           rz0 # Map the following over the lowercase input:
      /rz0d       #  Count occurrences of d in lowercase input
     t            #   minus 1
    !             #    inverted (0 -> True)
   &              #     and
           }dG    #      d is in the lowercase alphabet
.A                # Print whether all values are truthy

z=input()
print(all(map(lambda d:not z.lower().count(d)-1and d in "abcdefghijklmnopqrstuvwxyz",z.lower())))

C#, 82 bytes

bool f(string s)=>!!(s.GroupBy(c=>c).Any(c=>c.Count()>1|(!Char.IsLetter(c.Key))));

edit: added test for char

edit: using GroupBy to shorten it by 5 byte

APL (Dyalog Unicode), 25 20 22 bytes

''≡'(.).*\1|\d'⎕S'&'⍠1

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Returns 1 for true, else 0.

Saved 5 bytes thanks to @H.PWiz

Fixed, and saved another byte thanks to @Adám

How?

''≡'(.).*\1|\d'⎕S'&'⍠1 ⍝ Tacit fn
                    ⍠1 ⍝ Ignore case
               ⎕S'&'   ⍝ Search and return the match(es)
   '(.).*\1|\d'        ⍝ For this regex
''≡                    ⍝ And compare to the empty string

Tcl, 114 bytes

proc I w {lmap c [set L [split $w ""]] {if {[regexp -all -nocase $c $w]>1|![string is alp $c]} {return 0}}
expr 1}

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proc I w {lmap c [set L [split $w ""]] {if {[llength [lsearch -al -noc $L $c]]>1|![string is alp $c]} {return 0}}
expr 1}

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Still too long for my taste!